A magnet is suspended horizontally in the earth's magnetic field. When it is displaced and released, it oscillates in a horizontal plane with a period T. If a piece of wood of same moment of inertia as the magnet is attached along the length to the magnet, the new period of oscillation of the system would be -

\frac{T}{3}

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\frac{T}{2}

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\frac{T}{\sqrt{2}}

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\sqrt{2} T

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Solution

The correct option is D $\sqrt{2} T$
Due to wood, moment of inertia of the system becomes twice, but there is no change in magnetic moment of the system. So, here by using -
$T = 2 π \sqrt{\frac{I}{M B_{H}}}$ ....(1)
$T^{'} = 2 π \sqrt{\frac{2 I}{M B_{H}}}$ ....(2)
From equation (1) and (2),
$T^{'} = \sqrt{2} T$

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The correct option is D √2TDue to wood, moment of inertia of the system becomes twice, but there is no change in magnetic moment of the system. So, here by using - T=2π√IMBH ....(1) T′=2π√2IMBH ....(2) From equation (1) and (2), T′=√2T